Question 1

In how many ways, can 24 persons be seated around a circular table if there are is 13 seats?

Accepted Answer

First , we select 13 persons out of 24 persons in 24 C13 ways. Now, these 13 persons can be seated in 12! ways around a table . So required number of ways = 24 C13 x 12! = [24! / {13!(24 - 13)!}] x 12! = [24! / {13! x 11!}] x 12! = 24! / (13 x 11!)

Question 2

There are 14 points in a plane, out of which 4 are collinear. Find the number of triangles made by these points .

Accepted Answer

The required number of triangles = nC3 - mC3 Here, n = 14, m = 4 = 14C3 - 4C3 = (14 x 13 x 12 x 11! ) / (3! x 11!) - 4! / (3! x 1!) = (14 x 13 x 12)/6 - 4/1 = 14 x 26 - 4 = 364 - 4 = 360

Question 3

A question paper consists of two sections having respectively 3 and 5 questions. The following note is given on the paper. "It is not necessary to attempt all the question". One question from each section is compulsory. In how many ways, a candidate…

Accepted Answer

Here, we have two sections A and B. Section A has 3 question and B has 5 question and one question from each section is compulsory according to the given condition. ∴ Number of ways selecting one or more than one question from section A = 23 - 1 = 7 Similarly, from section B = 25 - 1 = 31 According to the rule of multiplication, the required number of ways in which a candidate can select the question = 7 x 31 = 217

Question 4

There are 10 stations on a railway line. The number of different journey tickets that are required by the authorities, is

Accepted Answer

There are 10 station on railway line. So, the number of different journey tickets between two station from given 10 stations from one side = 10C2 = 10 x 9/2 = 45. Similarly, number of different journey tickets from other side = 45 ∴ Total number of tickets to be generated by authorities. = 45 + 45 = 90

Question 5

The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a group consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Y is its member, is

Accepted Answer

If Mrs. X is selected among the ladies in the committee, then Mr. Y is not selected or if Mrs. X is not selected then Mr. Y can be there in the committee... So, required number of ways = 8C3 x 6C4 + 7C3 x 7C4 = [(8 x 7 x 6)/(3 x 2)] x [(6 x 5)/(2 x 1)] + [(7 x 6 x 5)/(3 x 2)] x [(7 x 6 x 5)/(3 x 2)] = 840 + 1225 = 2065

Question 6

A five-digit number divisible by 3 is to be formed using digits 0, 1, 2, 3, 4 and 5 without repetition, the total number of ways this can be done, is

Accepted Answer

A five-digit number, which is divisible by 3, is formed when sum of digits is also divisible by 3. So, combination formed using six-digits, which are divisible by 3 = 5 + 4 + 3 + 2 + 1 = 15 = 5 + 4 + 2 + 1 + 0 = 12 So, set of number are (5, 4, 3, 2, 1) and (5, 4, 2, 1, 0). Number formed by using 1st set = 5 x 4 x 3 x 2 x 1 = 120 Similarly, using 2nd set = 4 x 4 x 3 x 2 x 1 = 96 Hence, using 2nd set, underlined place cannot be filled by 0, otherwise it will become a four-digit number. ∴ Total number = 120 + 96 = 216

Question 7

A new flag is to be designed with six vertical stripes using some or all of the colour yellow, green,blue and red. Then, the number of ways this can be made such that no two adjacent stripes have the same colour is

Accepted Answer

Any of the 4 colour can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe). Similarly, There are 3 ways to colour each of the remaining stripes. ∴ The number of ways the flag can be coloured is 4(3)5 = (12) (3)4 = 12 x 81

Question 8

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2,....., 9 such that the first-digit of code is non-zero. The code, handwritten on a slip, can however potentially create confusion when read upside down, for examples th…

Accepted Answer

The available digits are 0, 1, 2,...., 9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digit repetition not allowed). Thus, the code can be made in 9 x 9 = 81 ways. Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways Total number of ways confusion can arise = 4 x 3 = 12 Thus, the ways in which no such confusion arise = 81-12 =69

Question 9

How many numbers can be formed from 1, 2, 3, 4, 5, ( without repetition), when the digit at the unit's place must be greater than that in the ten's place?

Accepted Answer

The digit in the unit's place should be greater than that in the tens' place. Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order,hence required number = 4! However, if digit 4 occupies the unit place then 5 cannot occupy the ten;s position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways. Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. this can happen in 2 ways. Hence, we will have (2 x 3! ) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers = 4! + (3) x 3! + (2 x 3!) + 3! = 4! + 6 x 3! = 24 + (6 x 6) = 60

Question 10

Boxes numbered 1, 2, 3, 4, and 5 are kept in a row and they are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of…

Accepted Answer

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4, and 5) with either blue or red balls 25 = 32. Two adjacent boxes with blue can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234)and (345). Four boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345). And five boxes with blue can be got in 1 way. Hence, the number of ways of filling the boxes such that adjacent boxes have blue = (4 + 3 + 2 + 1) = 10. Hence, the number or ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22.

Question 11

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Accepted Answer

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ( 7 C 3 * 4 C 2 ) = 210. Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 120 Required number of ways = (210 x 120) = 25200.

Question 12

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

Accepted Answer

The smallest number in the series is 1000, a 4-digit number. The largest number in the series is 4000, the only 4-digit number to start with 4. The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3. The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4. Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999. Including 4000, there will be 376 such numbers.

Question 13

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

Accepted Answer

When at least 2 women are included. The committee may consist of 3 women, 2 men : It can be done in 4 C 3 * 6 C 2 ways or, 4 women, 1 man : It can be done in 4 C 4 * 6 C 1 ways or, 2 women, 3 men : It can be done in 4 C 2 * 6 C 3 ways. Total number of ways of forming the committees = 4 C 2 * 6 C 3 + 4 C 3 * 6 C 2 + 4 C 4 * 6 C 1 = 6 x 20 + 4 x 15 + 1x 6 = 120 + 60 + 6 =186

Question 14

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

Accepted Answer

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways. If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways. If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary. The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Question 15

12 people at a party shake hands once with everyone else in the room.How many handshakes took place?

Accepted Answer

There are 12 people, so this is our n value. So, 12 C 2 1= 66

Question 16

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

Accepted Answer

The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Question 17

How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

Accepted Answer

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines. AB , AC BA , BC CA , CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

Question 18

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Accepted Answer

The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Question 19

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Accepted Answer

Required number of ways = ( 7 C 5 * 3 C 2 ) = ( 7 C 2 * 3 C 1 ) = 63

Question 20

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

Accepted Answer

A team of 6 members has to be selected from the 10 players. This can be done in 10 C 6 or 210 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210×6= 1260

Permutation and Combination Questions